Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x^2 - 6x}{x^3 + 4x^2 - 60x} \div \dfrac{3x + 15}{4x^2 + 32x - 80} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{x^2 - 6x}{x^3 + 4x^2 - 60x} \times \dfrac{4x^2 + 32x - 80}{3x + 15} $ First factor out any common factors. $n = \dfrac{x(x - 6)}{x(x^2 + 4x - 60)} \times \dfrac{4(x^2 + 8x - 20)}{3(x + 5)} $ Then factor the quadratic expressions. $n = \dfrac {x(x - 6)} {x(x + 10)(x - 6)} \times \dfrac {4(x + 10)(x - 2)} {3(x + 5)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {x(x - 6) \times 4(x + 10)(x - 2) } { x(x + 10)(x - 6) \times 3(x + 5)} $ $n = \dfrac {4x(x + 10)(x - 2)(x - 6)} {3x(x + 10)(x - 6)(x + 5)} $ Notice that $(x + 10)$ and $(x - 6)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {4x\cancel{(x + 10)}(x - 2)(x - 6)} {3x\cancel{(x + 10)}(x - 6)(x + 5)} $ We are dividing by $x + 10$ , so $x + 10 \neq 0$ Therefore, $x \neq -10$ $n = \dfrac {4x\cancel{(x + 10)}(x - 2)\cancel{(x - 6)}} {3x\cancel{(x + 10)}\cancel{(x - 6)}(x + 5)} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $n = \dfrac {4x(x - 2)} {3x(x + 5)} $ $ n = \dfrac{4(x - 2)}{3(x + 5)}; x \neq -10; x \neq 6 $